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Question
Solve `dy/dx=1+xy` with initial condition `x_0=0,y_0=0.2` By Taylors series method. Find the approximate value of y for x= 0.4(step size = 0.4).
Solution
The Taylor series is given by,
`y=y_0+xy'_0+x^2/(2!) y_0''+x^3/(3!) y_0''' ` …….. (1)
With `x_0=0,y_0=0.2,x=0.4`
Now , `y'=1+xy` ∴ `y'_0=1`
`y''=y+xy' ` ∴ `y_0''=y_0=0.2`
`y'''=y'+y'+xy''`
=`2y'+xy''`
=`2y'+xy''` ∴ `y_0'''=2y'_0=2`
`y''''=2y''+y''+xy'''` ∴` y_0''''=3y_0+xy_0'''=0.6`
Putting these values in equation 1, we get
`y=0.2+(0.4)1+(0.4)^2/(2!) 0.2 +(0.4)^3/(3!).2+(0.4)^4/(4!).(0.6)+........`
`y=0.2+0.4+0.016+0.02132+0.00064`
`y=0.63797`
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